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## #1 Re: This is Cool » sums of power sequences » 2017-08-25 03:34:48

almighty100 wrote:

Thank you both for the reply.  cmowla replied on his youtube video, as the links he was trying to post for me (to his work on the problem) apparently caused him to be banned... for a only few days, hopefully?

I guess the spam filter block time is up.

For those interested, I replied to almighty100 with this pdf in one the comments in the third video of my video series.  In the hyperlinked thread which contained my first online post about this topic, if you go both backward and forwards from that post, you can see other proofs of specific powers of i.  (For example, if you go ahead, you will find how I proved Sum i^1 by writing a linear equation from two "data points", and you can see a system of equations approach for calculating i^2 from someone else a few posts before that.)

## #2 Re: This is Cool » Something ineteresting » 2017-07-06 09:10:31

As I discovered, you can also modify Pascal's Triangle to help to construct the positive integer power sum formulas (image, video series) and the Bernoulli Numbers.

## #3 Re: This is Cool » A Cool Relationship between Permutations and Calculus! (Not Gamma Fn) » 2016-04-02 10:53:39

I forgot to mention this, but for those who don't have Mathematica, see !6 versus !6 (with this formula) to compute subfactorial.

You can substitute any integer in place of the 6's and get the correct subfactorial value.

## #4 This is Cool » A Cool Relationship between Permutations and Calculus! (Not Gamma Fn) » 2016-04-02 10:49:52

cmowla
Replies: 1

Besides the gamma function,

(an integral) being a link between permutations and calculus, I have found a derivative relationship for positive integers (with help, as I will explain shortly).  I'm not sure if it's a new formula or not (probably not), but I couldn't find it on the web anywhere, so I wanted to put it on the web!

Inputing in Mathematica, we can see that it is equivalent to n!

How I Found It

In the process of analyzing cycle types for a Rubik's cube theory project (of which I have probably mentioned it on here before), I wrote this post on speedsolving.com.

It's a long post, but just click the "spoiler" button (which we call "messages" on here) entitled "Formula Derivation".

I begin it with:

2 Pieces(k = 2)

3 Pieces(k = 3)

4 Pieces (k = 4)

5 Pieces (k = 5)

6 Pieces (k = 6)

7 Pieces (k = 7)

8 Pieces (k = 8)

9 Pieces (k = 9)

10 Pieces (k = 10)

11 Pieces (k = 11)

I had to find a way to express the general term of the sequence:

{1, 1, 3, 11, 53, 309, 2119, 16687, 148329, 1468457,...}  (Which is left fraction's numerator, "some #".)

Just inputting this sequence into Mathematica, I found this OEIS sequence page.

On that page you will find:

c = CoefficientList[Series[Exp[ -z]/(1 - z)^2, {z, 0, 30}], z] For[n = 0, n < 31, n++; Print[c[[n]]*(n - 1)! ]]

As it says right beneath this mathematica code, it was submitted by user Zerinvary Lajos in 2009.  Unfortunately, he died in 2011 (the year before I even made this post on speedsolving).  May he (continue to) rest in peace.

Continuing on,
If we type the above line in Mathematica, we can type the next line to get the desired sequence above.

c*Table[n!, {n, 0, 30}]

Of course, the coefficients of a Maclaurin series are found via taking the nth derivative, and thus that's where the nth derivative comes from in my formula.  For convenience, here's a link to Wolfram|Alpha obtaining the number 53 in the above sequence by inputting 6 for n.  Notice that I had to take the (k-2)th derivative to obtain the correct number in the sequence which corresponds to the above pattern for 6.

Therefore, we can express the

as (for k not equal to zero)

Since

then

The Mathematica input below illustrates that this expression gives the correct subfactorial values.

Finally, we need to take the summation of this expression.  Clearly k cannot be zero (as stated when we simplified), and when k is one, the entire sum is zero.  Therefore, we start with k=2.  (It also makes sense to start with k=2 because we are taking the (k-2)th derivative.)

Once we evaluate the summation for a few values of n, it should be clear that it is one less than the correct corresponding factorial value, and thus we add one to achieve the (my) formula which we wanted to derive.

## #5 Re: This is Cool » sums of power sequences » 2015-06-12 04:59:55

Yesterday I created and uploaded a three part video series on the content I have posted in this thread previously (with the exception of the Bernoulli number stuff).

I present the method in "Part 1", abstract it in "Part 2", and prove it in "Part 3".  Enjoy!

I realized that I made a few mistakes in the proof(s) in the messages/spoilers in this post, but my video proof is correct.  In addition, I have created my own proof of

Therefore my proof is now both complete and correct.

Let me know what you guys think!

## #6 Re: This is Cool » Formulas for the solution of Diophantine equations. » 2014-07-08 05:16:06

individ wrote:

The value of the method of calculation is much greater than the threat delete the message.

Then I don't understand why you post your solutions to begin with.

Do you think that it's *impossible* for someone to figure out how you find these solutions?  If not, then if your discovery means anything to you, you need to stop giving your results to people, because someone can come along and figure out your method and legally claim it as his own.  Even if you are a genius, there are other geniuses out there.  This is just algebra after all.

Your choice to post solutions without explaining a method to get them is going to aggravate anyone who cares about the subject matter, as you are just *teasing* them.  If doing this has become a highlight in your life, then by no means stop, but I am puzzled that you are in shock of the reactions you are getting.  In fact, I'm surprised they didn't ban you yet for being a *spammer*, because that is exactly how you are behaving as far as forums are concerned.

Forums are not personal blogs (which I see you have one).  Forums are a community, and the nature of a community is for its members to share resources with others to benefit all.  You're not doing that, and therefore the nature of your contributions, even though they might be great, are of no use to any forum.

For the same reason, I don't see the point of this thread on this *forum* other than being a personal release of satisfaction for yourself.  There's nothing wrong with that, but you can't expect anyone, even those who originally were interested, to do anything but ignore you after a short while, which I obviously see here.  I mean, what can we say?  If you are not willing to post the method, what do you want us to do?

Sure, you find a solution to one D equation which was unknown before and you post it.  Great.  Then you post solutions for dozens (if not hundreds) of D equations without an explanation.  If you sincerely feel that you are benefiting us or anyone else on any other forum by the nature of your posts, can you please tell us how?  I'm curious to know.

If you are aware of all of this, then I see no reason why you are complaining.

If you are trying to prove yourself on this forum, you've gotten our attention.  What's next?

## #7 Re: This is Cool » Formulas for the solution of Diophantine equations. » 2014-07-07 05:42:33

May I suggest to overwrite Google Translate's "Integers and sets us." at the end of *every* post you make here and on the math.stackexchange forum with "Integers set by us".

Perhaps that's why you are not being understood and not appreciated.  Miscommunication.

I also would suggest doing this in your second post of this thread, as I didn't understand what was going on until I realized that is what you meant to say.

## #8 Re: This is Cool » Prime Numbers!! » 2014-06-25 08:33:21

kerr wrote:

http://en.wikipedia.org/wiki/Largest_known_prime_number
How is 17,425,170 a prime number? It is divisible by 2.
Thanks

It clearly says in plain English that this number is the number of digits of the largest known prime number.  You're welcome?

## #9 Re: This is Cool » nxnxn Rubik's Cube Theory » 2014-06-23 07:34:19

Since the title of this thread is a general one, aside from my current project, I note that the following is an alternate formula which I found that calculates the number of positions of the nxnxn cube (I'm sure some of you all have seen it before using the modulo function):

And, it's derivative can be written as:

See my posts in this thread (and the thread other members from there linked to in which I also posted in) for further details, or, if anyone has questions, ask away.

Who would have thought prime numbers would be present in such formulas!  (Perhaps those who are actually interested in prime numbers might be able to interpret this in a way I cannot).

## #10 Re: This is Cool » I need a help » 2014-05-24 21:16:21

Stangerzv wrote:

n=0 is not a trivial because when n=0,

Okay, but besides my misuse of terminology, you can still use what I wrote.  Any comment (or "thank you") for the rest of what I wrote?

BTW, I don't know why you put this thread in the "This is Cool" sub-forum.  It should have been placed in the "Help Me!" sub-forum.

## #12 Re: This is Cool » nxnxn Rubik's Cube Theory » 2014-05-12 07:48:37

The following is one concept which comes up in my proof which I am not sure is new in abstract algebra.  It's obviously well-known that the 3x3x3 Rubik's cube (and the 2x2x2 Rubik's cube) can be studied using Group Theory, but...

Has anyone ever heard of a conditional identity element in a non-Abelian group?  The reason I ask is that for a specific portion of my proof, there are three identity elements which can be used for a certain procedure, two of which can only be used on a condition.  If we violate this condition, then we are no longer in the Rubik's cube group.  (Note that I don't use such terminology in the proof to describe this phenomena, but this one of the concepts shown concretely and in simple English).

I guess what I'm really asking is if such a concept is considered undergraduate math (I'm not sure because I haven't heard of this before).  This is one of the examples for which I had in mind when I said earlier

cmowla wrote:

Perhaps those who are fluent with graduate level topics related to the content in my proof can say the same thing I say in plain English with condensed symbols and technical language and interpret them in their relation to group theory, for example.

## #13 Re: This is Cool » nxnxn Rubik's Cube Theory » 2014-05-09 15:37:00

ShivamS wrote:

I am sure that your dissertation is interesting, but in the past 2 years I have never spoken to anyone submitting their dissertation which contains absolutely no graduate level mathematics.

I'm glad you mentioned this because this IS my main concern.  Now I say the following with all due respect:

Perhaps those who are fluent with graduate level topics related to the content in my proof can say the same thing I say in plain English with condensed symbols and technical language and interpret them in their relation to group theory, for example.  In addition, perhaps there exists knowledge which may shorten my proof significantly.

However, my primary goal was to make it as simple as possible to understand without doing anything fancy (that is, I purposely did not study any academic work which is related to this, if there even is such work, so that I could explain the proof to almost anyone who was interested in simple language) or using what I call "encrypted" language to say something simple in an overcomplicated way in my eyes and other laymen.  I also did this so that my proof would be completely independent of other previously published works.

ShivamS wrote:

1. In your second post in this thread, you used "believe" a lot. For example, you believe that you proved it. This leads me to believe that obviously your dissertation has not been peer reviewed. This means it could potentially contain fallacies. I suggest you post some parts that you think may be incorrect here.

I guess I shouldn't have used that term because I'm positive there are no fallacies, as I have tested my theory with actual Rubik's cube scrambles along the way, and I have at least triple checked everything.  I am very hesitate to say the word "positive", but, based on how I proved it, I left no room for error, especially since I didn't do anything fancy which can certainly cause more room for error.

ShivamS wrote:

I also highly suggest that you go to a local university and have your paper reviewed.

Even though I doubt (with good reason) that I have a fallacy, it would certainly be good to have at least two separate reviews to gain the confidence of the public.  This is certainly a necessity.

ShivamS wrote:

3. Unfortunately, a PhD by publication is often considered to be of less value then a regular PhD (I have a couple of sources), including to your employer (in academia and the industry).

I am aware of this, and it's completely understandable I suppose (I guess it really all depends on what traits your employer is looking for, what you have published to get a PhD, etc.).  I have not planned on getting a PhD to actually use it, but if I have done something which can technically earn me one anyway, why not I suppose, as I might be able to use it to advance myself somehow in the future when my life goals change.  In addition, as I mentioned already, if it is published as a dissertation, people who are familiar with graduate level mathematics will be interested to read it and possibly create unforeseen derivative works from it (who knows, I guess).

But please, if I have a misconception about any of this (which is very possible), please let me know.

ShivamS wrote:

To counteract all of these easily, have you considered enrolling in a regular PhD program, completing the coursework in one year and giving your thesis defence and getting the PhD? I think that is the best method for you.

That sounds nice, but is it only a year of course work?  Even though I have potential dissertation after only doing less than 2 years of research and work, I can't say that I could finish advanced course work at an accelerated pace (although I'm not positive, I'm almost certain that I just have an average IQ...the process of writing this proof made me feel as though I have a below average IQ to be honest).

Thanks for the advice.  I really appreciate your time, and I hope to hear from you again as well as others.

## #14 Re: This is Cool » nxnxn Rubik's Cube Theory » 2014-05-09 09:41:43

ShivamS wrote:

Obviously you have a thesis advisor who helped you develop this, so if he thinks this is okay then go on with it.

I actually don't have one, as I'm not in graduate school.  I've been doing this research as a hobby, and thus I was interested in PhD by publication, possibly.

I actually got the topic from Per Kristen Fredlund in a thread he started on a Rubik's cube forum.  (I have the same username there as I do here).  I had a little help from members qqwref and Stefan, but not that much.  Most of this was good old trial and error because I never had to do something quite like this for my undergraduate degree, and it seems like Rubik's cube theory so far has either been limited to pure group theory or pure computer brute force algorithms ("god's number is 20", for example), but I haven't heard too much about what can come from combining both (although I did not do a brute force search, as it wasn't necessary because the theory and brain power was a more natural substitute).

There are a few people in the world, at least, who are doing projects which involve both, but I don't believe any of these projects are as groundbreaking as this one (I'm not bragging, rather, I feel as if I am observing, not producing, this proof because the process of writing it and doing all calculations was a VERY humbling and out of this world experience), and thus I thought perhaps I should make it a dissertation so that it will be recognized by audiences who can possibly make derivative works from it which could lead to real world applications in the future.

Recently, however, I found out that at least 10 years ago, some people were wondering about this (for the 3x3x3 Rubik's cube, at least), and thus Per wasn't the first to pose the question online.  Link

## #15 Re: This is Cool » nxnxn Rubik's Cube Theory » 2014-05-09 06:22:45

Agnishom wrote:

If you are asking if it is worthy of a PhD topic, it most certainly is.

my proof is certainly long enough

Mathematicians like simplicity. They do not judge a proof by its length. In fact, the shorter the proof the more beautiful it is.

Anyhow, if you have truly done something like that, very good. Well Done!

Wow, I thought it might be, but I wasn't positive.  My proof is mainly proof by exhaustion, and thus, although lengthy, I kept it very easy to understand and as compact as I could without making it more complicated than it must be.  The same goes for the code and its documentation.  (I just programmed functions in Mathematica, nothing fancy).

Thanks for your opinion, and if you would like to post a any 3x3x3 scramble (I can do those quickly, but I could solve up to a 7x7x7 cube scramble if you like...that's as big as I will go, however) to see the results, feel free to post any scramble, and I will give a solution for it and link it to an online cube simulator so that you don't have to execute it yourself (the solutions are quite lengthy).

## #16 Re: This is Cool » nxnxn Rubik's Cube Theory » 2014-05-09 05:56:16

Agnishom wrote:

Where is the proof?

It's just on my PC (I didn't post it on the web).  So, assuming that my proof is correct, would the result I proved be enough, or is the content of the proof the main issue with PhD dissertations (my proof is certainly long enough)?

## #17 Re: This is Cool » nxnxn Rubik's Cube Theory » 2014-05-09 05:07:37

(For those who have seen this thread before, I edited the first post to simply be a more intense problem on the same subject matter.  Much has changed since then as well.)

I believe I have proved the answer to this question to be YES.

I believe that I have not only proved it, but I created a systematic human method to solve the nxnxn cube explicitly on paper with abstract solutions which then are translated into actual Rubik's cube moves.

If you don't know what the first post is mentioning or you would just like to see an example, post a 3x3x3 Rubik's cube scramble (with some known notation so that I can interpret it), and I will post a solution to it using my method.

If my proof is indeed correct, do any of you mathematicians (I have a BS in math myself) think it's worthy to be a PhD dissertation topic in some branch of pure mathematics?  I didn't do any research in previous existing math libraries to gather any information (or used any previously existing results of proof) to write this proof, and thus I'm not sure which branch of math this falls under exactly.  Certainly this topic has never been done before.

The proof expands on basic facts about permutations, perhaps some calculus (for the nxnxn cube in general), programming (in Mathematica), and perhaps some well known results from abstract algebra (which are "common talk" in the "cubing community").  My goal was to make this proof as simple as possible so that everyone could understand it if they devoted the time to just read it, and thus I'm beginning to wonder if it is "complex" enough to be a dissertation topic (perhaps it's easy for me to understand because I have been working on this proof for a year and a half).

The amount of new knowledge it brings is great (but I don't see any real world applications of it besides solving the Rubik's cube without reducing it into different subgroups...that is, my method-which is used to test my theory-can be used to solve all pieces of the nxnxn cube in one step), and thus that is why I thought I would kindly ask for some opinions from you guys (as well as other people).

## #18 Re: This is Cool » sums of power sequences » 2013-03-29 08:41:56

Stangerzv wrote:

Hi cmowla

I happened to read the article regarding the sums of power of integer using Bernoulli & Pascal long time ago. Could look similar but you can read it here www.sanjosemathcircle.org/handouts/2008-2009/20081112.pdf

When I was in the process of deriving the formula which I used to create the "adjusted pascal triangle" for the sum of power formulas, I actually went through the same process that Bernoulli did for a portion of my research.

Leaving out all of the ugly trial and error that I did (it was helpful to have a CAS handy through it all), here is my full derivation of my formula

which I used to create that "adjusted Pascal's Triangle" for the power sum polynomial formulas.

[Step 1]:  I derived a formula for sum i^2 from scratch.

We need to show that

[Step 2]:  Extrapolate

By applying the same pattern

to different power sums, I found that a general formula for sum i^R is simply:

, which has been known for a while, but I didn't know that unfortunately until after I found it.

This is the only step which is not fully justified, should one consider this entire derivation to possibly be a single proof, and is why I said I derived it "from scratch" (emphasizing with the quotes that there was one point of intuition).  However, one could simply find a proof for that formula, since it has existed for a while, and combine it with the remaining to have a complete proof that my "Adjusted Pascal's Triangle" is indeed correct to construct the power sum polynomials for positive integers.

[Step 3] Make a recurrence relation.

The above general formula isn't useful unless we can somehow use it to calculate the power sum sequence for the next power using all previous formulas.  The following is the recurrence relation I came up with which proved to be useful.  If the expression in the summation of the formula below is evaluated at any positive integer R>0, then the greatest exponent of i is R-1, i.e., i^(R-1), which is precisely what we need in a recurrence relation.

So we need to show that:

[Step 4]: Transforming into a pattern which can be generated by a diagram (which turned out to be an "adjusted Pascal Triangle")

This was the trickiest part.  I probably spent more time doing this step than anything else because I didn't even know if it was possible to create a formula which could generate an intuitive diagram.

So we show that

## #19 Re: This is Cool » Exploring mathematicians and smart mathematicians » 2013-03-16 06:18:07

I think I'm more of the experimental type myself.  I guess "smart" could mean conceptual and "exploring" could mean experimental.  Being an experimentalist by nature, I had a very difficult time with upper level math courses in college such as advanced calculus and abstract algebra before I got my B.S. in math.  Now for abstract algebra at least, I'm working on a Rubik's cube project which I believe is teaching me a thing or two about the real idea behind abstract algebra that I was aware of but didn't see any application for.

I am sorry to say that I can't comment on your proof (because I honestly don't care enough about calculus at the moment to have interest), but a little while back I too looked at a topic from advanced calculus which I wanted to construct a proof which I could understand better for some of the basic limit laws.  (What do you think about those?  Are they valid?).  Hopefully someone will comment on your proof, I was certainly lucky to get at least one response on mine.

I think proofs which explain more and show every step of the mind are more elegant than compact proofs which assume the reader has "mathematical maturity".

I am many things, but incapable of understanding complicated concepts is not one of them (even though I still debate that sometimes for no good reason).  For example, I created this "parity algorithm" for the 4x4x4 Rubik's cube from scratch.  I explained the entire process that went through my mind, different paths I tried before, and the logic behind the entire process in a video about an hour and a half long (in 4 parts).
Part I.
The reason why that algorithm is so significant is not just because it scrambles the cube so much (see the beginning of this video to see what a
"normal" parity algorithm
looks like.  Rather, the minimum number of 90 degree slice turns to "flip an edge" before I came around was 25 (like that "normal" algorithm has).  Mine uses only 19...

One of the lines of bobbym's signature says that "90% of mathematicians do not understand 90% of currently published mathematics."  I wonder why that is?  Hmmm?  [1] I do realize that there is so much information published now that it would be impossible for any mathematician to be able to understand a significant portion of published topics because he/she could not possibly read it all in his or her lifetime.  But I am assuming that bobbym's quote is saying that, if a mathematician is presented with published math at random, he could not understand what he reads 9 times out of 10. [2] However, I theorize that the main problem is poor communication.  That is, when most mathematicians write their papers, they assume that their readers understand every elementary concept which is prerequisite knowledge to whatever it is they are writing about.  There are so many ways to say the same thing, and we have to be careful how to write something down in technical language, especially if people's lives depend on it, but I still don't like how, say, the language in which the majority of wikipedia's math articles are written in, for example.  I thought taking discreet math would have helped me to understand proofs because I learned what the basic symbols mean, but even if we assume that a particular proof does not build upon a proven theorem or lemma, one word or term in a proof that the reader isn't familiar with might require that the reader must read another topic to become familiar with that word, and this continues...I have come to many dead ends that I cannot possibly understand what someone is trying to say because even the prerequisite knowledge is written in a language that I cannot understand nor care to spend a lifetime trying to learn.

## #20 Re: Help Me ! » probability question help me » 2013-03-15 19:57:19

This problem doesn't make sense.

lucik1900 wrote:

You are holding a party at home, and everyone is about to participate in the following game.
Each person will write their name on a card. All the cards will then be collected and randomly redistributed (one per person).

Think of the "randomly distributed" deck of cards be a random permutation scramble of n objects, where n is the number of people you have in your group.  If no one gets a card with his or her own name on it, then this is a derangement.

The number of possible ways in which a deck of cards can be distributed to the party members at random is n!.  The number of derangement distributions/scrambles is subfactorial of n or !n.

Now this is where the problem doesn't make sense to me.

lucik1900 wrote:

When everyone has a card with someone else's name on it, you will call out the name on your card.

Okay, so this is telling me that only when there is a derangement will any names will be called out.

lucik1900 wrote:

If anyones name does not get called out at some stage during this game, they will have to drink a whole 1 litre bottle of vodka by midnight.

The phrasing of this is odd because by the quoted text above this one, no one's name will be called when there isn't a derangement.  So when the first round occurs where at least someone has his or her own card, then no body's name will be called and therefore everyone will have to drink before midnight (is this question a joke or an encrypted message saying that everyone is going to a drink vodka before midnight at a party?).

Since we weren't given a specific number of rounds the game has, and since the name card distributions are done at random, we can definitely assume that when enough rounds are executed, everyone will have to drink before midnight.

So no matter what, every round that "counts" towards determining the fate of any given players blood alcohol levels is one in which everyone will have to drink vodka.

So all I can say is the probability that no one person gets back his or her card in a single distribution round is subfactorial(n)/n!, which is a general formula which holds true for any n.

So if we were given that there were 15 rounds in the game, and we had say, 5 players, then the probability that 0 persons will not have gotten back his or own card in any of those 15 rounds would be

which pretty much guarantees that all people will have to drink before midnight.

BarandaMan wrote:

They are equal, that was very useful for eliminating A, so we could make it 2A on the LHS, really thank you for the trick.

You're welcome.  Go look at the post again.  I realized I used the wrong code for phi, but I corrected it now.

BarandaMan wrote:

Really appreciate the replies; but obviously a tad confused because two people have two different answers

I corrected a mistake in my work (I wrote a beta in the denominator instead of a gamma) when I substituted equ3b into equ4b, and I now have the same answer as bob bundy.

BarandaMan wrote:

On cmowla's post, why does the denominator (1-theta) just change to -(theta-1) ??

Because they are equal (just multiply in that negative one).  This is useful to do sometimes for algebraic simplification.

bob bundy wrote:

hi cmowla

Sorry.  I'm not getting the same as you.

Bob

Is there a mistake in my work?  I provided all steps.

EDIT:
I found the mistake.  It's in the last step before "step 3".  I'll fix it and see if I got the same answer as you.

I solved for A, elliminating A*,e,r, treating all symbols as variables.  Note that I capitalized a to A to differentiate it between alpha a little more.  In fact, you will find from the work below that alpha does not even make up A at all.

I provided all steps I took so that you can double check my work (I don't solve systems of equations that often anymore).

Problem:
Solve for A in terms of anything but A*,e, and r using the following system of equations:

Solution:

EDIT: I corrected the mistake and got the correct answer.